Question: From the following infinite list of numbers, how many are integers? $$\sqrt{4096},\sqrt[3]{4096},\sqrt[4]{4096},\sqrt[5]{4096},\sqrt[6]{4096},\ldots$$
Solution: Since $4096=2^{12}$, one of these is an integer if the number on the root is a factor of 12. Therefore, the only numbers in the list that are integers are $\sqrt{4096}=2^6=64$, $\sqrt[3]{4096}=2^4=16$, $\sqrt[4]{4096}=2^3=8$, $\sqrt[6]{4096}=2^2=4$ and $\sqrt[12]{4096}=2$. This makes $\boxed{5}$ integers in all.